no, I am combining like terms
call\[(3^{20})=x\]we then have\[(3^{20})+(3^{20})+(3^{20})=x+x+x=3x\]sub back in what we know x is...\[3x=3(3^{20})\]now we use the property\[x^a\cdot x^b=x^{a+b}\]and get\[3(3^{20})=3^1\cdot3^{20}=3^{20+1}=3^{21} \]